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# MOLES - PowerPoint PPT Presentation

MOLES. A collection of objects. Mole:. A collection of Avogadro’s number of objects. Avogadro’s number. 6.022 * 10 23. Atoms Particles Ions Cations Anions Molecules Formula Units. Mole Ratio.

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Presentation Transcript

A collection of objects

Mole:

A collection of Avogadro’s number of objects

6.022 * 1023

Atoms

Particles

Ions

Cations

Anions

Molecules

Formula

Units

The coefficient in front of each component in a balanced equation

2 Al (s) + 6 HCl (aq) ----- 2 AlCl 3(aq) + 3 H2(g)

2 moles of Al reacts w/ 6 moles HCl

to produce

2 moles AlCl3 and 3 moles H2 gas

This states:

Argon

atoms

1 mole

Helium

atoms

MOLE CALCULATIONS

One mole of argon atoms ….

….. is a number

One mole of helium atoms ….

….. is a number

6.022 * 1023 Argon atoms

6.022 * 1023 Helium atoms

….. has a mass

….. has a mass

39.9 grams

4.00 grams

Elements on p.table = I mol

Molar mass (mass/1 mol)

1. Elements

- decimal # on p.table

Ti 47.87 amu or g

2. Molecules/Cmpd

- sum of mass of all elements

mass = 1 mole

Divide

by

Multiply

by

MASS

How many moles are in 345.6 g NaNO3 ?

Step 1: find formula wt. of NaNO3

1 Na = 23.0 g

1 N = 14.0 g

3 O = 3 * 16.0 = 48.0 g

1 mole of NaNO3 = 85.0 g

Step 2 : Find # of moles

Use factor-label method

4.07 moles NaNO3

How many grams are in 0.6 moles N2O ?

Step 1: find molecular wt. of N2O

2 N = 2 * 14.0 = 28.0 g

1 O = 16.0 g

1 mole of N2O = 44.0 g

Step 2 : Find mass

Use factor-label method

26.4 g N2O

by

Divide

by

MOLES

6.02*1023

NUMBER of

PARTICLES

How many molecules are in 1.6 moles oxygen?

Step 1: Recognize that mass does not apply here.

Step 2 : Find # molecules

Use factor-label method

9.63 * 1023 molecules O2

How many moles are in 3.01 *1012 atoms of Chromium?

Step 1: Recognize that mass does not apply here.

Step 2 : Find # moles

Use factor-label method

5.00 * 10-12 mols Cr

by

Divide

by

MOLES

6.02*1023

mass = 1 mole

Divide

by

Multiply

by

MASS

NUMBER of

PARTICLES

There is not a direct relationship

between MASS and PARTICLES

Will need to use MASS at some

point in the problem

Given # formula units, find grams

How many grams are in 1.208 * 1024 formula units AgCl ?

Step 1: find formula wt. of AgCl

1 Ag = 107.9 g

1 Cl = 35.5 g

1 mole of AgCl = 143.4 g

Step 2 : Find moles

Step 3 : Find mass

Use factor-label method

288 g AgCl

Converts moles

to MASS

Converts formula units

to MOLES

Will need to use MASS at some

point in the problem

Given grams, find # atoms

How many atoms are in 128.0 grams of Mercury ?

Step 1: find molar mass of Hg

1 mole of Hg = 200.6 g

Step 2 : Find moles

Step 3 : Find atoms

Use factor-label method

3.84*1023 atoms Hg

Converts mass

to MOLES

Converts moles

to ATOMS

PROBLEM

What is the formula weight of sodium carbonate,

Na2CO3. This is an industrial chemical used in

making glass.

Also, equivalent to 1mole

of a substance

SOLUTION

2 * 23.0 = 46.0

2 Na

1 C

3 O

1 * 12.0 = 12.0

3 * 16.0 = 48.0

106.0 g

“STOICHIOMETRY”

Sodium Phosphate (aq) + Barium Nitrate (aq) ------

Barium Phosphate (s) + Sodium Nitrate (aq)

2

Na3PO4(aq) + Ba(NO3)2(aq) ----- Ba3(PO4)2(s) + NaNO3 (aq)

3

3

6

Aluminum (s) + Hydrochloric Acid (aq) ---------->

Aluminum Chloride (aq) + Hydrogen (g)

3

Al + HCl ---- AlCl3 + H2

2

6

2

3

(amts of 2 reactants given: limiting reactants)

Solid calcium metal burns in air (oxygen) to form a calcium oxide cmpd.

Using 4.20 g Ca & 2.80 g O2 how much pdt is made?

Find 1) grams of pdt made from g of Ca

2) grams of pdt made from g of O2

3) limiting reagent

4) how much pdt can be made

5) % yield if 3.85 g produced

PLAN: Need ------> balanced chem. rxn

STEPS:

g Ca -----> mols Ca -----> mols pdt ------> g pdt

M g Ca coeff X pdt/Y react M g pdt

(same steps converting oxygen)

2 2

Ca(s) + O2(g) -------> CaO(s)

M g Ca M g O2 coeff X pdt/Y react M g pdt

40.1 g Ca:CaO 2:2 56.1 g

32.0 g O2:CaO 1:2

5.89 g

9.82 g

2 Ca(s) + O2(g) -------> 2 CaO(s)

M g : 40.1g 56.1 g

Coeff : 2 1 2

mols :

0.105

0.175

0.105

0.0875

3) limiting reagent:

Which produced the least amt-- Ca or O2?

Ca ---> CaO = 5.89 g O2 ---> CaO = 9.82 g

Ca, limiting reagent

5.89 g CaO

4) how much pdt can be made:

5) % yield=

RECAP: -- Balanced chem. equation --

then,GIVENUSEFIND

grams Molar Mass mols

mols coeff mols

mols Molar Mass grams

limiting reagent: the reactant that produces the least

amt of moles or mass of a specific pdt